The energy needed for Li(g) ------> Li3+(g) + 3e is 1.96 * 10^4 J/mol. If the first ionisation enthalpy of Li is 520kJ/mol. Calculate the second ionisation enthalpy of Li.
Dear student,
Li(g)--------> Li+3(g) + 3e-............. 1.96 x 104 kJ/mol = 19600 kJ/mol
Li(g)--------> Li+(g) + e-.................. 520 kJ/mol
Li+2 is a single electron system, hence its ionisation energy can be given by bohr model
Ionisation energy of Li+2 = 13.6 x 9 = 11793 kJ/mol
Second ionisation energy = Total energy - IE1 - IE3 = 19600 -520-11793 = 7287 kJ/mol
Please see, all energies are in kJ/mol including total energy.
Hope it helps
Regards
Li(g)--------> Li+3(g) + 3e-............. 1.96 x 104 kJ/mol = 19600 kJ/mol
Li(g)--------> Li+(g) + e-.................. 520 kJ/mol
Li+2 is a single electron system, hence its ionisation energy can be given by bohr model
Ionisation energy of Li+2 = 13.6 x 9 = 11793 kJ/mol
Second ionisation energy = Total energy - IE1 - IE3 = 19600 -520-11793 = 7287 kJ/mol
Please see, all energies are in kJ/mol including total energy.
Hope it helps
Regards